Find roots of polynomial $$ p(x) = x^3+13x^2+40x $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -5\\[1 em]x_3 &= -8 \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x }$ from $ x^3+13x^2+40x $ and solve two separate equations:

$$ \begin{aligned} x^3+13x^2+40x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^2+13x+40 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^2+13x+40 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.

Step 2:

The solutions of $ x^2+13x+40 = 0 $ are: $ x = -8 ~ \text{and} ~ x = -5$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

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